Re: Einstein's lopsided caricature of space and time


"Thomas Smid" <thomas.smid@xxxxxxxxx> wrote in message news:1125145354.441513.273350@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Dirk Van de moortel wrote:
> > "Thomas Smid" <thomas.smid@xxxxxxxxx> wrote in message news:1125088821.431078.3940@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> > >
> > > Dirk Van de moortel wrote:
> > > > "Thomas Smid" <thomas.smid@xxxxxxxxx> wrote in message news:1125072903.307356.115390@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> > > > >
> > > > > Dirk Van de moortel wrote:
> > > > >
> >
> > [snip]
> >
> > > > > > Okay... next points.
> > > > > >
> > > > > > 18) Whatever the value or meaning of x', according to you,
> > > > > > the observer, in order for the distance (as measured
> > > > > > by you, the observer) between an object with velocity v,
> > > > > > and a light signal with velocity c to grow from 0 to x',
> > > > > > will take a time x'/(c-v), as measured in your frame.
> > > > > >
> > > > > > 19) Whatever the value or meaning of x', according to you,
> > > > > > the observer, in order for the distance (as measured
> > > > > > by you, the observer) between an object with velocity v,
> > > > > > and a light signal with velocity -c to shrink from x' to 0, it
> > > > > > will take a time x'/(c+v), as measured in your frame.
> > > > > >
> > > >
> > > > I'll suppose that 18 and 19 were OK.
> > > >
> > > > > > 20) Whatever the value or meaning of x', according to you,
> > > > > > the observer, a light signal going away from an object (that
> > > > > > has velocity v with respect to you), and returning to it after
> > > > > > having been reflected by some mirror that, at the event of
> > > > > > reflection, is at a distance x' (according to you, sitting in
> > > > > > your chair), takes the time x'/(c-v) + x'/(c+v)
> > > > >
> > > > > Can you elaborate? Where is the mirror and is it co-moving with the
> > > > > object or the chair?
> > > >
> > > > The movenment of the mirror doesn't really matter.
> > > > The only function it has, is that it reflects the signal.
> > > > It could be co-moving with the object, or with you,
> > > > or with anyone. Or it could be non-existent immediately
> > > > before it reflects, and then explode right after it has
> > > > reflected the signal, and done its job.
> > > > The only thing that matters is that the mirror happens
> > > > to be present at a distance (as measured by you) x'
> > > > from the object when it (according to you) it reflects
> > > > the signal.
> > > > This way, you can apply points 18 and 19 to find the
> > > > flight times of the signal before and after the reflection
> > > > event.
> > > >
> > > > So to fix something, let's say that it comoves with the
> > > > object and that it always has a distance x' (as measured
> > > > by you of course!) between itself and the object.
> > > > Then you could say that the object and the mirror are
> > > > in fact tied to some rod with length x' (according to you),
> > > > moving away from you at velocity v. The object would
> > > > then be attached to the rear end of the rod (let's call that
> > > > point A), and the mirror would be attached to the front
> > > > end of the rod (let's call that point B).
> > > >
> > > > This way, at time t on your clock, since the object is at
> > > > distance v t to you, the mirror is at distance v t + x',
> > > > according to you. If you like, you can call this number
> > > > x, such that x = v t + x' is the coordinate of the mirror
> > > > at time t.
> > > >
> > > > But none of the above is actually essential.
> > >
> > > Why would it not be essential? If the object and mirror would be moving
> > > relatively to each other, the distance x' between them would not be the
> > > same anymore when the signal is reflected, so you could not use
> > > x'/(c-v) +x'/(c+v) as the total travel time.
> >
> > This seems to be an important point.
> >
> > The distance would be defined to be x' at the precise
> > moment of reflection. What does the mirror do?
> > It receives a signal and is supposed to instantaneously
> > reflect it. Actually, it does not even matter if there is a
> > mirror at all. There could be just as well an explosion
> > that is triggered by the arrival of the outgoing signal on
> > a detonator. Then the part of the light of the explosion
> > that comes toward you would play the part of the
> > reflected signal.
> > It wouldn't make any difference: the speed of the signal
> > is the same, and the distance covered to the object would
> > be x' (as seen by you in your frame).
> > What matters is the single event of reflection, namely
> > the birth of a signal heading toward the object and
> > yourself, sitting in that chair.
> > Our basic assumption is still valid: the speed of some
> > light signal that comes to you is independent of the
> > motion (if any) of what produced it.
> >
> > Let's look again at point 19, with which you agreed:
> >
> > 19) Whatever the value or meaning of x', according to you,
> > the observer, in order for the distance (as measured
> > by you, the observer) between an object with velocity v,
> > and a light signal with velocity -c to shrink from x' to 0, it
> > will take a time x'/(c+v), as measured in your frame.
>
> I don't question this. But as mentioned, you are stricty speaking not
> dealing with the distance between the signal and the object here
> between the corresponding flags you are setting in your system.

Well, we are strictly speaking dealing with the distance that
you, sitting in your chair, measure or calculate between
A) the point where an object is at a certain time, and
B) the point where a signal is at that same time.
(remark, time, as measured on your clock again).
We arbitrarily give this distance the name x'. But that is
just a name. We could just as well call it "George".

I think we implicitly agreed that we can just as well say that
we are (still strictly speaking) dealing with the distance that
you, sitting in your chair, measure or calculate between
A) an object at a certain time, and
B) a signal at that same time
(remark, time, as measured on your clock again).
You know the velocities of the object and the signal, with
respect to yourself, so you can perfectly calculate when
the will meet, according to you.

OK?
If yes, we continue,
Otherwise, we must keep dwelling on this point... Just stop
me here again then.
But I will make an exception at this point, and continue with
your next comment. [Please, next time try to stop after the first
objection - otherwise, this quickly gets unmanageable]

>
> Apart from this, I noticed that your time including the reflection
> x'/(c-v)+ x'/(c+v) is actually also wrong if the distance between the
> object and the mirror is constant, even when you are not dealing with a
> light signal but usual particles:

As I said, we don't need a mirror. We could even have the
signal sort of "change its mind" at a certain point, and then
start heading toward the object. All we *essentially* need,
is to deal with the distance as agreed upon (by you) at some
certain time (on your clock), between an object and a signal,
where the distance of that signal does a transition from
A) growing from 0 to x' as seen by you, and
B) shrinking from x' to 0 as seen by you.
We had agreed in points 18 and 19 that the first part
takes a time x'/(c-v), whereas the second part takes a
time x'(c+v).
So obviously, still according to you, sitting in your chair,
the total time for the signal to go from the object and to
return to it, must be the sum of these times.

OK?
If not, what exactly should that total time be?

I will snip the remainder of your comment, because you
talk about quantities as measured in the moving frame
- you used the phrase "velocity U (in A's and B's frame)"
We clearly are not ready for that.
We might get there as soon as we completely agree over
everything that is defined, measured and calculated in your
frame only, you sitting in your chair.
Let's take one step at a time.

Dirk Vdm


.

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